Chapter 3-98 - uu;llljj PDF

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Chapter 3

Torsion

3.1 Introduction Torsion : twisting of a structural member, when it is loaded by couples that produce rotation about its longitudinal axis T1

=

P1 d1

the couples torques,

T2 T1,

twisting

=

T2

couples

P2 d2 are called

or

twisting

moments unit of

T

:

N-m,

lb-ft

in this chapter, we will develop formulas for the stresses and deformations produced in circular bars subjected to torsion, such as drive shafts, thin-walled members analysis of more complicated shapes required more advanced method then those presented here this chapter cover several additional topics related to torsion, such statically indeterminate members, strain energy, thin-walled tube of noncircular section, stress concentration, and nonlinear behavior

3.2 Torsional Deformation of a Circular Bar consider a bar or shaft of circular cross section twisted by a couple

T,

assume the left-hand end is fixed and the right-hand end will rotate a small angle

,

called angle of twist

1

if every cross section has the same radius and subjected to the same torque, the angle

(x)

will vary linearly between ends

under twisting deformation, it is assumed 1. plane section remains plane 2. radii remaining straight and the cross sections remaining plane and circular 3. if



is small, neither the length

consider an element of the bar small element

dx,

L

nor its radius will change

on its outer surface we choose an

abcd,

during twisting the element rotate a small angle a state of pure shear, and deformed into max

=

b b' CC ab

=

r d CC dx

2

ab'c'd,

d,

the element is in

its shear strain

max

is

d / dx

represents the rate of change of the angle of twist

 = d / dx

, denote

as the angle of twist per unit length or the rate of twist, then

max in general, pure torsion,

r

= 

and





is constant along the length (every cross section is

are function of

x,

in the special case of

subjected to the same torque) 

=



then

C

max

=

r CC

L

L

and the shear strain inside the bar can be obtained 

=

 =

 C

max

r for a circular tube, it can be obtained min =

r1 C

max

r2 the above relationships are based only upon geometric concepts, they are valid for a circular bar of any material, elastic or inelastic, linear or nonlinear

3.3 Circular Bars of Linearly Elastic Materials shear stress

 in the bar of a

linear elastic material is 

=

G

G : shear modulus of elasticity

3

with the geometric relation of max 

Gr

=

=

the shear strain, it is obtained

G

=

C 

max

r



and



in circular bar vary linear with the radial distance 

from

the center, the maximum values max and max occur at the outer surface the shear stress acting on the plane of the cross section are accompanied by shear stresses of the same magnitude acting on longitudinal plane of the bar if the material is weaker in shear on longitudinal plane than on cross-sectional planes, as in the case of a circular bar made of wood, the first crack due to twisting will appear on the surface in longitudinal direction a rectangular element with sides at 45 o to the axis of the shaft will be subjected to tensile and compressive stresses

The Torsion Formula consider a bar subjected to pure torsion, the shear force acting on an element is

 dA,

the moment of this force about

the axis of bar is dM

dA

=

  dA

  dA

4

equation of moment equilibrium = ∫ dM

T

A

= G  Ip in which

=

A

G  ]

=

= ∫ 2 dA

Ip

= G ∫ 2 dA

A

[

is the polar moment of inertia

A

 r4 Ip

2 =∫ G   dA

= ∫   dA

A

 d4 =

CC

for circular cross section

CC

2

32

the above relation can be written 

T =

CC

G Ip G Ip : torsional rigidity 

the angle of twist 

=

L

can be expressed as TL

=



CC

is measured in radians

G Ip torsional flexibility torsional stiffness

f k

= =

L CC

G Ip G Ip CC

L and the shear stress is 

=

G

=

T G  CC G Ip

the maximum shear stress

max

T

=

CC

Ip at

5



=

r

is

max

=

Tr CC

16 T

=

CC

 d3

Ip for a circular tube

=  (r24 - r14) / 2 =  (d24 - d14) / 32

Ip

if the hollow tube is very thin  (r22 + r12) (r2 + r1) (r2 - r1) / 2

j

Ip

 (2r2) (2r) (t)

=

=

2  r3 t

=

 d3 t / 4

limitations 1. bar have circular cross section (either solid or hollow) 2. material is linear elastic note that the above equations cannot be used for bars of noncircular shapes, because their cross sections do not remain plane and their maximum stresses are not located at the farthest distances from the midpoint

Example 3-1 a solid bar of circular cross section d = 40 mm,

L = 1.3 m,

G = 80 GPa

(a) T = 340 N-m,

max,

(b) all = 42 MPa,

all = 2.5o,

(a)

max

=

16 T CC

=

=

3

=

=

 d4 / 32 TL CC

G Ip

=

T=?

16 x 340 N-M CCCCCCC

=

27.1 MPa

3

d Ip

?

 (0.04 m) =

2.51 x 10-7 m4 340 N-m x 1.3 m

CCCCCCCCCC

-7

80 GPa x 2.51 x 10 m 6

4

=

0.02198 rad = 1.26o

all

(b) due to

= 42 MPa

T1 =  d3 all / 16 =  (0.04 m)3 x 42 MPa / 16 = 528 N-m all

due to

2.5o

=

=

G Ip all / L

=

674 N-m

T2

thus

Tall

=

=

2.5 x  rad / 180o

=

0.04363 rad

80 GPa x 2.51 x 10-7 m4 x 0.04363 / 1.3 m

=

min [T1, T2]

=

528 N-m

Example 3-2 a steel shaft of either solid bar or circular tube T all

= =

1200 N-m,

all

=

40 MPa

0.75o / m

G

=

78 GPa

(a) determine

d0

of the solid bar

(b) for the hollow shaft, t = d2 / 10, determine d2 (c) determine

d2 / d0,

Whollow / Wsolid

(a) for the solid shaft, due to d03

= 16 T /  all

d0

=

due to Ip

all

=

40 MPa

= 16 x 1200 /  40

0.0535 m =

= 152.8 x 10-6 m3

53.5 mm

0.75o / m = 0.75 x  rad / 180o / m = 0.01309 rad / m

all =

= T / G all = 1200 / 78 x 109 x 0.01309 = 117.5 x 10-8 m4

d04

= 32 Ip / 

d0

=

= 32 x 117.5 x 10-8 / 

0.0588 m

thus, we choose

=

= 1197 x 10-8 m4

58.8 mm

d0 = 58.8 mm

[in practical design, d0 = 60 mm]

(b) for the hollow shaft d1

=

d2

-

2t

= d2

7

-

0.2 d2

=

0.8 d2

all

due to Ip

=

d23

=

d2

=

= -6

258.8 x 10 m 0.0637 m

all

due to

40 MPa

0.05796 d24

=

= 0.05796 d24

=  [d24 - (0.8d2)4] / 32

=  (d24 - d14) / 32

Ip

T r /all

=

63.7 mm

0.75o / m

=

all

= 0.01309

d2 4

=

d2

= 0.0671 m

= 1200 (d2/2) / 40

3

=

0.01309 rad / m = 1200 / 78 x 109 x 0.05796 d24

= T / G Ip

2028 x 10-8 m4

thus, we choose

= 67.1 mm

d0 = 67.1 mm

[in practical design, d0 = 70 mm]

(c) the ratios of hollow and solid bar are d2 / d0

=

CCC

Whollow

67.1 / 58.8

CCC

Ahollow

=

Wsolid

= =

Asolid

1.14

CCCCCC  (d22 - d12)/4  d02/4

=

0.47

the hollow shaft has 14% greater in diameter but 53% less in weight

Example 3-3 a hollow shaft and a solid shaft has same material, same length, same outer radius and

ri

=

R,

0.6 R for the hollow shaft

(a) for same T, compare their , , and W (b) determine the strength-to-weight ratio (a)

∵ 

= T R / Ip

∴ the ratio of (Ip)H

=



 R2 /2

or -



=



is the ratio of

T L / G Ip

 (0.6R)2 /2

8

=

1 / Ip

0.4352  R2

(Ip)S

 R2 /2

=

0.5  R2

=

(Ip)S / (Ip)H

=

0.5 / 0.4352

thus

1

=

H / S

=

(Ip)S / (Ip)H

=

1.15

also

2

=

H / S

=

(Ip)S / (Ip)H

=

1.15

3

= WH / WS = AH / AS =  [R2 - (0.6R)2] /  R2

the hollow shaft has

=

15%

1.15

greater in



and

,

= 0.64 but

decrease in weight (b) strength-to-weight ratio TH

= max Ip / R

TS

=

WH

SH

is

=

max (0.5  R4) / R

=

0.64  R2 L 

WS

SH

=

TH / WH

SS

=

TS / WS

36%

Tall / W

= max (0.4352  R4) / R

max Ip / R

=

thus

S

=  R2 L 

0.5 max R /  L

=

SS

3.4 Nonuniform Torsion (1) constant torque through each segment = - T1

-

TBC

=

- T2



=

- T1 n

 i

i=1

=

T2

n



+

T3

TAB

=

T L CC

i=1

i

=

= 0.68 max R /  L

greater than

TCD

= 0.4352  R3 max

i

Gi Ipi

(2) constant torque with continuously varying cross section

9

- T1

0.5  R3 max

36%

d

T dx CCC

=

G Ip(x) L



L

∫ d

=

∫

=

0

0

T dx CCC

G Ip(x)

(3) continuously varying cross section and continuously varying torque L



∫

=

0

CCC

T(x) dx

L

∫ d

=

0

G Ip(x)

Example 3-4 a solid steel shaft ABCDE,

d

=

30 mm

T1

=

275 N-m T2

=

450 N-m

T3

=

175 N-m G

=

80 GPa

L1

=

500 mm L2

=

400 mm

max

determine

-

T1

TCD

= T2

TBC

=

- T1

=

CCC

BC

=

16 TBC 3

=

=

=

=

BC

CD

3

d

BD

= Ip

=

+

CC  d4 32

CCCCCC

=

51.9 MPa

CCCCCC

=

33 MPa

16 x 275 x 103 3

 30

CCC 16 TCD

175 N-m

- 275 N-m

d

CD

BD

in each part and

16 x 175 x 103 3

 30

=

CCC  304 32

10

=

79,520 mm2

BC

CCC

=

=

CCC

=

=

0.011 rad

80 x 103 x 79,520

= - 0.0216 + 0.011 = - 0.0106 rad = - 0.61o

BC + CD

=

CCCCCCCC 175 x 103 x 400

G Ip

BD

- 0.0216 rad

80 x 10 x 79,520

TCD L2

=

=

3

G Ip

CD

CCCCCCCC - 275 x 103 x 500

TBC L1

Example 3-5 a tapered bar

AB

of solid circular

cross section is twisted by torque d = dA

at A,

d = dB

T

at B,

determine

max

(a)

constant over the length,

(b)

T

=

thus

max

max

=

and 

≧ dA

dB

of the bar

occurs at

dmin [end A]

16 T CCC  dA3

angle of twist d(x)

=

dA

+

d -d CCC x B

A

L

Ip(x)

=

CC  d4

C (d

=

32

then 

L

= ∫

0

+

A

32

d -d CCC x) B

A

4

L

T dx 32 T dx CCC = CC ∫ CCCCCCC G I (x) G d -d (d + CCC x) L

0

p

B

A

A

L

to evaluate the integral, we note that it is of the form

11

4

dx ∫CCCC (a + bx)4 if we choose of

 

a

1 - CCCCC 3 b (a + bx)3

=

=

dA

and

b

= (dB - dA) / L,

then the integral

can be obtained =

32 T L

1 ( CC 3G(dB - dA) dA3

1

-

CCCCCC

CC

dB3

)

a convenient form can be written 

=

TL

2 +  + 1

CCC

( CCCCC ) 3 3

G IpA where



=

dB / dA

IpA

=

 dA4 / 32

in the special case of a prismatic bar,

 = 1,

3.5 Stresses and Strains in Pure Shear for a circular bar subjected to torsion, shear stresses act over the cross sections and on longitudinal planes an stress element

abcd

is cut

between two cross sections and between two longitudinal planes, this element is in a state of pure shear we now cut from the plane stress element to a wedge-shaped element, denote A0

the area of the vertical side face, then

the area of the bottom face is A0 tan ,

12

then



= T L / G Ip

and the area of the inclined face is

A0

sec  summing forces in the direction of  A0 sec  

or

=

=

 A0 sin  +

 A0 sec  





 A0 tan  cos 

2  sin  cos  =

summing forces in the direction of

or



=

=  (cos2

 sin 2 

 A0 cos  sin2)

-

 A0 tan  sin  =



is plotted in figure

()max = 

at

 = 0o

()min = - 

at

 = ! 90o

()max = ! 

at

 = ! 45o

and

vary with

the state of pure shear stress is equivalent to equal tensile and compressive stresses on an element rotation through an angle of 45o if a twisted bar is made of material that is weaker in tension than in shear, failure will occur in tension along a helix inclined at 45o, such as chalk

Strains in pure shear if the material is linearly elastic 

 cos 2

=

/G

13

where

G

is the shear modulus of elasticity

consider the strains that occur in an element oriented at  = 45o, applied at 45o

min = - 

and



=

applied at

 = - 45

 = 45o

then at max

CC

max

=

-

E

at

max

 = - 45o



CC

=

- max

=